A technician attaches one lead of a digital voltmeter to the ground terminal of the TP sensor and the other meter lead to the ne
gative terminal of the battery. The ignition is turned on, engine off, and the meter displays 37.3 mV. Technician A says this is the signal voltage and is a little low. Technician B says the TP sensor ground circuit has excessive resistance. Who is right?
KE=1/2mv^2 - equation for kinetic energy KE=(1/2)(0.12 kg)((7.8 m/s)^2 - plug it into the formula KE=(0.06 kg)(60.84 m/s) - multiply 1/2 to the mass and square the speed KE= 3.7 J - answer Hope this helps
